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IAL 2026 Jan Q8

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 8

题目

Problem

(a) Given that u=xyu=xy, where uu is a function of xx (x>0)(x>0), show that

d2ydx2=1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac1x \left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) \end{align*}
(3)

(b) Hence show that the transformation u=xyu=xy transforms the differential equation

xd2ydx2+2(2x+1)dydx+13xy=17e3x−4y\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2(2x+1)\frac{\mathrm{d}y}{\mathrm{d}x} +13xy = 17\mathrm{e}^{3x}-4y \end{align*}

into the differential equation

d2udx2+4dudx+13u=17e3x\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\frac{\mathrm{d}u}{\mathrm{d}x} +13u = 17\mathrm{e}^{3x} \end{align*}
(2)

(c) Solve differential equation (II) to determine a general solution for uu in terms of xx.

(5)

(d) Hence determine the general solution of differential equation (I).

(1)

解答

(a)

解法一

思路

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从 u=xyu=xy 求导两次。第一次会得到 u′=xy′+yu'=xy'+y，第二次再整理，就可以把 y′′y'' 表示出来。

答题过程

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Since

u=xy,\begin{align*} u=xy, \end{align*}

differentiating gives

dudx=xdydx+y\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} = x\frac{\mathrm{d}y}{\mathrm{d}x}+y \end{align*}

Differentiate again:

d2udx2=xd2ydx2+dydx+dydx=xd2ydx2+2dydx\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} ={}& x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +\frac{\mathrm{d}y}{\mathrm{d}x} +\frac{\mathrm{d}y}{\mathrm{d}x}\\[3mm] ={}& x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} +2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Therefore

xd2ydx2=d2udx2−2dydx\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} ={}& \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

and since x>0x>0,

d2ydx2=1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac1x \left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) \end{align*}

解法二

思路

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因为 x>0x>0，我们可以将变换式变形为 y=uxy=\dfrac{u}{x}。然后对 xx 连续求导两次，得到 y′′y'' 關於 uu 及其导数的表达式，最后通过将已知关系代回，证明其与目标等式右边（RHS）完全等价。

答题过程

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Since x>0x>0, we can express yy in terms of uu and xx as:

y=  ux=  ux−1\begin{align*} y =&\,\, \frac{u}{x}\\[3mm] =&\,\, ux^{-1} \end{align*}

Differentiate with respect to xx using the Product Rule:

dydx=  x−1dudx−ux−2=  1xdudx−ux2\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\, x^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} - ux^{-2}\\[4mm] =&\,\, \frac1x \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{u}{x^2} \end{align*}

Differentiate again with respect to xx:

d2ydx2=  ddx(x−1dudx−ux−2)=  (x−1d2udx2−x−2dudx)−(x−2dudx−2ux−3)=  1xd2udx2−2x2dudx+2ux3\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\, \frac{\mathrm{d}}{\mathrm{d}x}\left( x^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} - ux^{-2} \right)\\[4mm] =&\,\, \left( x^{-1}\frac{\mathrm{d}^2u}{\mathrm{d}x^2} - x^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} \right) - \left( x^{-2}\frac{\mathrm{d}u}{\mathrm{d}x} - 2ux^{-3} \right)\\[4mm] =&\,\, \frac1x \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - \frac2{x^2} \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2u}{x^3} \end{align*}

Now examine the RHS of the target equation:

RHS=  1x(d2udx2−2dydx)=  1x[d2udx2−2(1xdudx−ux2)]=  1xd2udx2−2x2dudx+2ux3\begin{align*} \text{RHS} =&\,\, \frac1x \left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - 2\frac{\mathrm{d}y}{\mathrm{d}x} \right)\\[4mm] =&\,\, \frac1x \left[ \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - 2\left( \frac1x \frac{\mathrm{d}u}{\mathrm{d}x} - \frac{u}{x^2} \right) \right]\\[4mm] =&\,\, \frac1x \frac{\mathrm{d}^2u}{\mathrm{d}x^2} - \frac2{x^2} \frac{\mathrm{d}u}{\mathrm{d}x} + \frac{2u}{x^3} \end{align*}

Since LHS=d2ydx2=RHS\text{LHS} = \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \text{RHS}, the identity is proven:

d2ydx2=  1x(d2udx2−2dydx)\begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} =&\,\, \frac1x \left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) \end{align*}

(b)

解法一

思路

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把 (a) 的 y′′y'' 代入原方程，同时由 u′=xy′+yu'=xy'+y 得到 xy′=u′−yxy'=u'-y。这样所有含 yy 的项会抵消，留下关于 uu 的二阶方程。

答题过程

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From part (a),

xd2ydx2=d2udx2−2dydx\begin{align*} x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}

Also,

dudx=xdydx+y⟹xdydx=dudx−y\begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} = x\frac{\mathrm{d}y}{\mathrm{d}x}+y \quad\Longrightarrow\quad x\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x}-y \end{align*}

Substitute these into (I):

(d2udx2−2dydx)+2(2x+1)dydx+13u =17e3x−4y\begin{align*} &\left( \frac{\mathrm{d}^2u}{\mathrm{d}x^2} -2\frac{\mathrm{d}y}{\mathrm{d}x} \right) +2(2x+1)\frac{\mathrm{d}y}{\mathrm{d}x} +13u\\[3mm] &\,\hspace{2pt}=17\mathrm{e}^{3x}-4y \end{align*}

So

d2udx2+4xdydx+13u=17e3x−4yd2udx2+4(dudx−y)+13u=17e3x−4y\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4x\frac{\mathrm{d}y}{\mathrm{d}x} +13u ={}& 17\mathrm{e}^{3x}-4y\\[3mm] \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\left( \frac{\mathrm{d}u}{\mathrm{d}x}-y \right) +13u ={}& 17\mathrm{e}^{3x}-4y \end{align*}

The −4y-4y terms cancel, giving

d2udx2+4dudx+13u=17e3x\begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}x^2} +4\frac{\mathrm{d}u}{\mathrm{d}x} +13u = 17\mathrm{e}^{3x} \end{align*}

(c)

解法一

思路

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这是常系数非齐次二阶微分方程。先解辅助方程得到互补函数，再设特解为 λe3x\lambda\mathrm{e}^{3x}。

答题过程

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Solve the homogeneous equation using

m2+4m+13=0\begin{align*} m^2+4m+13=0 \end{align*}

Then

m=−4±16−522=−2±3i\begin{align*} m ={}& \frac{-4\pm\sqrt{16-52}}{2}\\[3mm] ={}& -2\pm3\mathrm{i} \end{align*}

So the complementary function is

uc=e−2x(Acos⁡3x+Bsin⁡3x)\begin{align*} u_c = \mathrm{e}^{-2x} (A\cos3x+B\sin3x) \end{align*}

For a particular integral, try

up=λe3x\begin{align*} u_p=\lambda\mathrm{e}^{3x} \end{align*}

Then

dupdx=3λe3x,d2updx2=9λe3x\begin{align*} \frac{\mathrm{d}u_p}{\mathrm{d}x} = 3\lambda\mathrm{e}^{3x}, \qquad \frac{\mathrm{d}^2u_p}{\mathrm{d}x^2} = 9\lambda\mathrm{e}^{3x} \end{align*}

Substitute into the differential equation:

9λe3x+4(3λe3x)+13λe3x=17e3x34λ=17λ=12\begin{align*} 9\lambda\mathrm{e}^{3x} +4(3\lambda\mathrm{e}^{3x}) +13\lambda\mathrm{e}^{3x} ={}& 17\mathrm{e}^{3x}\\[3mm] 34\lambda ={}& 17\\[3mm] \lambda ={}& \frac12 \end{align*}

Therefore the general solution for uu is

u=e−2x(Acos⁡3x+Bsin⁡3x)+12e3x\begin{align*} u = \mathrm{e}^{-2x}(A\cos3x+B\sin3x) +\frac12\mathrm{e}^{3x} \end{align*}

(d)

解法一

思路

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最后只需要用 u=xyu=xy，所以 y=uxy=\dfrac{u}{x}。

答题过程

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Since

u=xy,\begin{align*} u=xy, \end{align*}

we have

y=ux\begin{align*} y=\frac{u}{x} \end{align*}

Therefore

y=1x[e−2x(Acos⁡3x+Bsin⁡3x)+12e3x]\begin{align*} y = \frac1x \left[ \mathrm{e}^{-2x}(A\cos3x+B\sin3x) +\frac12\mathrm{e}^{3x} \right] \end{align*}